# How do you graph y=ln(x-k)+2?

Sep 1, 2017

First, note that the domain of $\ln \left(x\right)$ is $x > 0$. Then, the domain of $\ln \left(x - k\right)$ is $x - k > 0$, or $x > k$. We can neglect the $+ 2$ in determining this because it has no bearing on domain--it constitutes a vertical shift.

• Domain: $x > k$

Furthermore, note that $\ln \left(x\right)$ is asymptotic at $x = 0$. Symbolically, ${\lim}_{x \rightarrow {0}^{+}} \ln \left(x\right) = - \infty$. Here, this becomes:

• ${\lim}_{x \rightarrow {k}^{+}} \ln \left(x - k\right) = - \infty$

One other valuable thing to note is that $\ln \left(x\right) = 0$ at $x = 1$. Then, $\ln \left(x - k\right) = 0$ at $x - k = 1$, or at $x = k + 1$.

• $\ln \left(x - k\right)$ has a zero at $x = k + 1$.

Assuming we know the general shape of the logarithmic curve (which for $\ln \left(x\right)$ is a reflection of the exponential curve ${e}^{x}$ across the line $y = x$), we can draw the function $\ln \left(x - k\right)$.

Arbitrarily setting $k = 5$ for the sake of graphing the function in the Socratic graphing system, we know that $\ln \left(x - 5\right)$ will have the domain $x > 5$, have an asymptote to $- \infty$ at $x = 5$ and have a zero at $x = 5 + 1 = 6$. This looks like:

$\ln \left(x - 5\right)$

graph{ln(x-5) [1.01, 21.01, -6, 4]}

The missing piece here is that the function we want is $\ln \left(x - k\right) + 2$. So, all we do is we take each point on the graph of $\ln \left(x - k\right)$ and shift it up $2$. Again using our test function with $k = 5$:

$\ln \left(x - 5\right) + 2$

graph{ln(x-5)+2 [2.76, 22.76, -4.28, 5.72]}

If we're unsatisfied with this process of shifting vertically, we can also find the formal point at which $\ln \left(x - k\right) + 2$ has a zero. This is when $\ln \left(x - k\right) = - 2$, so $x = k + {e}^{-} 2$.

When $k = 5$, this translates to a zero at $x \approx 5.1353$.